f(x)=sinx-cosx=√2(√2/2sinx-√2/2cosx)=√2(sinxcosπ/4-cosxsinπ/4)=√2sin(x-π/4)
(1)因为-1≤sin(x-π/4)≤1,所以-√2≤f(x)≤√2,即f(x)的值域是[-√2,√2].
(2)当π/2+2kπ≤x-π/4≤3π/2+2kπ,即3π/4+2kπ≤x≤7π/4+2kπ(k∈Z)时,f(x)单调递减,
因此f(x)的单调递减区间是[3π/4+2kπ,7π/4+2kπ](k∈Z)；
由x-π/4=π/2+kπ,得x=3π/4+kπ(k∈Z)是f(x)的对称轴方程.